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How can I prove that $p:S^1\rightarrow S^1$, $z\mapsto z^2$ is a covering map? Please help. I was not able to prove it by applying definition of covering space.

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up vote 3 down vote accepted

Here's a different approach using some differential topology. If $f:M\to N$ is a surjective smooth map between manifolds and $df$ is everywhere an isomorphism, then $f$ is a covering map.*

Here, $z\mapsto z^2$ is a surjective map of one-manifolds and the differential is multiplication by $2$, in particular an isomorphism of each tangent space onto its image. Hence the map is a cover.

*The proof uses the inverse function theorem and is not difficult, so I'll leave it to you as an exercise. You can find it in standard differential topology textbooks like Guillemin-Pollack.

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Nice. Besides leaving the proof of the fact that proves what the OP asked, can you perhaps give some links to sites/books where this result can be found? – DonAntonio Feb 28 '13 at 12:00
@DonAntonio Edited as per request. Guillemin-Pollack is (I think) a standard reference for this sort of thing. – Neal Feb 28 '13 at 15:28

You can prove it by checking that the map is continuous, surjective, and that the 4 open semicircles are evenly covered.

enter image description here

LaTeX code for picture:

    \draw #1 circle (1);
    \draw[ultra thick,red,(-)] #1 ++ (#2:1) arc (#2:#2+180:1);
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Nice picture! How did you draw it? – user27126 Feb 28 '13 at 5:49
@Sanchez: Thanks! I added the LaTeX code I used to generate it. – Zev Chonoles Feb 28 '13 at 6:21
Thats really awsome! – Un Chien Andalou Feb 28 '13 at 6:22

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